Dependence of chemical potential to zero point of energy

Question

The chemical potential is defined as:
$$
mu = -Tfrac{ partial{S(N,V,E)} }{ partial{N} }
$$
It seems to me that this is completely independent of where I put the reference point of energy, because only the difference of entropy is relevant (and also temperature is defined as a difference in entropy).
However the Fermi-Dirac distrbution is:
$$langle n_r rangle = frac{1}{exp( beta(epsilon_r-mu) )+1}$$
But if I change the value of the reference point of energy, the value of $langle n_r rangle$ changes, which causes a contradiction. In my book about statistical mechanics, they state though that $epsilon_r-mu$ is independent of the reference point of energy, but I see not why, because $mu$ seems to be independent, while $epsilon_r$ doesn't seem independent of the reference point.

Oliver · Accepted Answer

Good question.  I think that you're incorrect to assume the chemical potential is independent of the zero point.

I think it's easiest to see that the chemical potential must change by using $mu = (partial U/partial N)_{S,V}$.  Suppose I first define the zero point as 0 so that the energy is $U(N)$.  Then I redefine the zero-point energy (for a particle) as $epsilon_0$, so that $U' = U + Nepsilon_0$.  Now,

begin{equation}
  mu' = left(frac{partial U'}{partial N}right)_{S',V} = left(frac{partial U}{partial N}right)_{S,V} + epsilon_0 = mu + epsilon_0
end{equation}

Then if $epsilon_r' = epsilon_r + epsilon_0$, you have that $epsilon_r' - mu' = epsilon_r - mu$, i.e. the difference is independent of the zero point.

In fact, the formula $mu = -T(partial S/partial N)_{U,V}$ doesn't imply that chemical potential is independent of zero point.  If you change the zero-point energy, you must adjust the form of $S$ accordingly.  For instance, you need to have $S(U=0) = S'(U' = Nepsilon_0)$ and so you see that $S'$ is different that $S$.

Dependence of chemical potential to zero point of energy

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